Question

Prove that $2\sqrt{3}$ is an irrational number.

Hint:

Multiplying an irrational number by a non-zero rational number always results in an irrational number.

Answer 1

Step 1: Assume the opposite (for contradiction).
Let $2\sqrt{3}$ be rational. Then we can write: \[ 2\sqrt{3} = \frac{p}{q}, \; \text{where } p \text{ and } q \text{ are integers, and } q \neq 0 \] Step 2: Simplify.
\[ \sqrt{3} = \frac{p}{2q} \] Step 3: Analyze.
Since $p$ and $q$ are integers, $\frac{p}{2q}$ is rational. Therefore, $\sqrt{3}$ would be rational.
Step 4: Contradiction.
But we know that $\sqrt{3}$ is irrational. Hence, our assumption is wrong.
Step 5: Conclusion.
Therefore, $2\sqrt{3}$ is an irrational number.